If 5 times 5th term is equal to the 6times the 6th term of an arithmetic sequence,then its 11th is?.
To find the 11th term of an arithmetic sequence given that 5 times the 5th term is equal to 6 times the 6th term, we need to determine the common difference (d) first.
Let's assume the first term (a1) of the arithmetic sequence is "a" and the common difference is "d".
The formula to find the nth term of an arithmetic sequence is:
an = a1 + (n-1)d
According to the given information:
5 * (a + 4d) = 6 * (a + 5d)
Now we can solve this equation for "d".
5(a + 4d) = 6(a + 5d)
5a + 20d = 6a + 30d
20d - 30d = 6a - 5a
-10d = a
So the common difference (d) is equal to -1/10 times the first term (a).
Now we can find the 11th term (a11) by substituting the values into the formula.
a11 = a1 + (11-1)d
a11 = a + 10d
Substituting -1/10 times a for d:
a11 = a + 10(-1/10)a
a11 = a - a
a11 = 0
Therefore, the 11th term of the arithmetic sequence is 0.
To find the 11th term of an arithmetic sequence given the information that 5 times the 5th term is equal to 6 times the 6th term, we need to use the formulas for the nth term and the common difference of an arithmetic sequence.
Let's start by understanding the formulas:
The nth term (Tn) of an arithmetic sequence is given by the formula:
Tn = a + (n - 1)d
where:
- Tn is the nth term of the sequence
- a is the first term of the sequence
- n is the position of the term in the sequence
- d is the common difference between consecutive terms
Now, let's use this formula to find the equation using the information provided:
For the 5th term:
T5 = a + (5 - 1)d
For the 6th term:
T6 = a + (6 - 1)d
Given that 5 times the 5th term is equal to 6 times the 6th term, we can write this as an equation:
5 T5 = 6 T6
Substituting the expressions for T5 and T6:
5 (a + 4d) = 6 (a + 5d)
Now, we can simplify the equation by expanding and combining like terms:
5a + 20d = 6a + 30d
Next, we can rearrange the equation to isolate one variable:
20d - 30d = 6a - 5a
-10d = a
Now that we have the value of 'a', we can find the 11th term by substituting the values into the formula for the nth term:
T11 = a + (11 - 1)d
T11 = -10d + 10d
T11 = 0d
Therefore, the 11th term of the arithmetic sequence is 0.
5(a+4d) = 6(a+5d)
a = -10d
T11 = a+10d = 0
check:
a=10, d=-1
10,9,8,7,6,5
5*6 = 6*5
a=-5, d=1/2
-5,-4.5,-4,-3.5,-3,-2.5
5(-3) = 6(-2.5) = -15
In both cases, T11 = 0